= Solution
The density is an equal mixture of the uniform distributions on $[0,1]$ and $[0,\theta]$. Hence
$$
\mathbb E_\theta X
=\frac12\cdot\frac12+\frac12\cdot\frac\theta2
=\frac{1+\theta}{4},
$$
so
$$
\boxed{\mathbb E_\theta(4\overline X-1)=\theta}.
$$
Thus $\widetilde\theta$ is unbiased.
Also,
$$
\mathbb E_\theta X^2=\frac{1+\theta^2}{6},
\qquad
\operatorname{Var}_\theta X
=\frac{5\theta^2-6\theta+5}{48}.
$$
The <central limit theorem> gives
$$
\sqrt n(\widetilde\theta-\theta)
\xrightarrow{d}
N\left(0,\frac{5\theta^2-6\theta+5}{3}\right).
$$
Replacing $\theta$ in the asymptotic standard error by the consistent estimator $\widetilde\theta$ gives the <Wald confidence interval>
$$
\boxed{
\widetilde\theta
\mathbin{\pm}
z_{1-\alpha/2}
\sqrt{\frac{5\widetilde\theta^2-6\widetilde\theta+5}{3n}}}.
$$
It may be intersected with the parameter space $(0,1)$.
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